Question

$\sqrt{3 \: } cosec140 - sec140 = 4$
prove that: sqrt(3) cosec140-sec140=4​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\sqrt{3}cosec140^{\circ}-sec140^{\circ}=4$

How To Prove :-

By taking L.H.S and we need to convert 'cosec and sec' ratios in the terms of 'sin and cos' ratios and we need to simplify it by multiplying and dividing the fraction by '2' so the value doesn't change. After by using some trigonometry formulas and we need to substitute value of some angles in that and we need to show that.

Formula Required :-

1) cosecA = 1/sinA

2) secA = 1/cosA

3) sin2A = 2sinAcosA

4) sin(A - B) = sinAcosB - cosAsinB

Solution :-

Taking L.H.S :-

$=\sqrt{3}cosec140^{\circ}-sec140^{\circ}$

$=\sqrt{3}\times\dfrac{1}{sin140^{\circ}}-\dfrac{1}{cos140^{\circ}}$

[ ∴ cosecA = 1/sinA

    secA = 1/cosA ]

$=\dfrac{\sqrt{3}}{sin140^{\circ}}-\dfrac{1}{cos140^{\circ}}$

Taking L.C.M :-

$=\dfrac{\sqrt{3}\times cos140^{\circ}-1(sin140^{\circ})}{sin140^{\circ}\times cos140^{\circ}}$

$=\dfrac{\sqrt{3}cos140^{\circ}-sin140^{\circ}}{sin140^{\circ} cos140^{\circ}}$

Multiplying and dividing with '2' on both sides :-

$=\dfrac{\sqrt{3}cos140^{\circ}-sin140^{\circ}}{sin140^{\circ} cos140^{\circ}}\times \dfrac{2}{2}$

$=\dfrac{2(\sqrt{3}cos140^{\circ}-sin140^{\circ})}{2sin140^{\circ} cos140^{\circ}}$

$=\dfrac{2(\sqrt{3}cos140^{\circ}-sin140^{\circ})}{ sin2(140^{\circ})}$

[ ∴ sin2A = 2sinAcosA ]

$=\dfrac{2\sqrt{3}cos140^{\circ}-2sin140^{\circ}}{sin280^{\circ}}$

Taking '4' as common in numerator :-

$=\dfrac{4\left(\dfrac{2\sqrt{3}}{4}cos140^{\circ}-\dfrac{1}{4}\times 2sin140^{\circ}\right)}{sin280^{\circ}}$

$=\dfrac{4\left(\dfrac{\sqrt{3}}{2}cos140^{\circ}-\dfrac{1}{2}sin140^{\circ}\right)}{sin280^{\circ}}$

Finding value of 'cos420°' and 'sin420°' :-

cos420° = cos(450 - 30)

= sin30°

= 1/2

∴ cos420° = 1/2

sin420° = sin(450 - 30)

= cos30°

= √3/2

∴ sin420° = √3/2

Substituting these values in the above equation :-

$=\dfrac{4\left(sin240^{\circ}cos140^{\circ}-cos240^{\circ} sin140^{\circ}\right)}{sin280^{\circ}}$

$=\dfrac{4[sin(420^{\circ}-140^{\circ})]}{sin280^{\circ}}$

[ ∴ sin(A - B) = sinAcosB - cosAsinB ]

$=\dfrac{4[sin420^{\circ}]}{sin420^{\circ}}$

= 4

= R.H.S

Hence Proved.